Conditions about $A$ for $\|Ax\|_2 \leq \|x\|_2$ to be true

Question

Suppose $A$ is a projection. What are the conditions that $A$ needs to satisfy for

$\|Ax\|_2 \leq \|x\|_2$ to hold.

Two possibilities I can see are:

• $A$ is an orthogonal projection.

• The largest singular value of $A$ is less than or equal to 1.

Are there any other specific conditions on $A$ which satisfy this?

Answer 1

Your statement implies that the space must be contracting. Since $A$ can be decomposed via SVD, then the only necessary condition for this, I believe, is that the largest singular value be less than or equal to 1.

Answer 2

HINT:

Recall that $A$ is a projection. Let $u\in \textrm{Im} A$, $v \in \textrm{Ker} A$. We have $A(u+\epsilon v) = u$. Therefore $$\|u+ \epsilon v \|\ge \|u\|$$ for all $\epsilon \in \mathbb{R} $ (or $\mathbb{C}$).

This implies $\langle u, v\rangle = 0$.

We conclude that $A$ must be an orthogonal projection.

Answer 3

As you stated $\lVert Ax \rVert_2 \leq \lVert x \rVert_2, \forall x \iff \bar{\sigma}(A) \leq 1$ where $\bar{\sigma}$ denotes the largest singular value. If $A$ is orthogonal then $\lVert Ax \rVert_2 = \lVert x \rVert_2, \forall x$. These conditions have nothing to do with $A$ being a projection and applies to all matrices.